ha lvh262 Homework 13 1 karakurt 56295 This print out should have 16 questions Multiple choice questions may continue on the next column or page find all choices before answering 001 10 0 points Find the domain of the vector function r t h t3 t 2 ln 9 t i 1 2 t 9 1 1 limit h 0 8 0 i 2 limit h 6 0 4 i 3 limit h 0 0 0 i 4 limit h 6 0 4 i 5 limit h 6 8 4 i 6 limit h 6 8 0 i correct Explanation For a vector function 2 2 t 9 correct 3 2 t 9 r t h f t g t h t i 4 t 2 t 9 the limit 5 2 t 9 lim r t Explanation A vector function t 0 r t h f t g t h t i is defined when each of f t g t and h t is defined Now for the given function is defined only when t 2 On the other hand h t ln 9 t is defined only 9 t 0 Consequently the domain of r t consists of all t E lim f t lim g t lim h t t 0 t 0 t 0 But for the given vector function lim f t lim 6 cos t 6 f t t3 is defined for all t while g t t 2 D t 0 t 0 while lim g t lim 8et 8 t 0 t 0 and lim h t lim t ln t 0 t 0 t 0 using L Hospital s Rule Consequently 2 t 9 lim r t h 6 8 0 i t 0 keywords vector function domain power function square root function log function 002 10 0 points keywords vector function limit trig function log function exponential function Find lim r t when t 0 003 t r t h 6 cos t 8e 4t ln t i 10 0 points ha lvh262 Homework 13 1 karakurt 56295 A space curve is shown in black on the surface 2 satisfies the equations y t 2 x t 2 z t 2 z y t 0 keywords surface space curve vector function 3D graph circular cylinder 004 y 10 0 points A space curve is shown in black on the surface z x Which one of the following vector functions has this space curve as its graph x 1 r t correct h t cos t t t sin t i y t 0 2 r t h sin 2t cos t sin t i 3 r t h cos t sin t sin 2t i 4 r t h cos t sin t sin 4t i 5 r t h sin 4t cos t sin t i Which one of the following vector functions has this space curve as its graph 6 r t h t cos t t sin t t i t 0 1 r t h sin t cos t t i 7 r t h t cos t t sin t t i 2 r t h cos t sin t t i 8 r t h t cos t t t sin t i 3 r t h cos t sin t sin 4t i correct Explanation The surface is a cone such that vertical cross sections perpendicular to y axis intersect the surface in a circle On the other hand the space curve exists only for y 0 But of the given vector functions only r t h t cos t t t sin t i t 0 4 r t h sin t cos t cos 2t i 5 r t h cos t sin t cos 4t i 6 r t h cos t sin t cos 2t i Explanation If we write r t h x t y t z t i ha lvh262 Homework 13 1 karakurt 56295 then 2 2 x t y t 1 for all the given vector functions showing that their graph will always lie on the cylindrical cylinder x2 y 2 1 To determine which particular vector function has the given graph we have to look more closely at the graph itself Notice that the graph oscillates with period 4 so r t is one of h cos t sin t sin 4t i h cos t sin t cos 4t i On the other hand it passes it through 1 0 0 and 0 1 0 Consequently the space curve is the graph of r t h cos t sin t sin 4t i 3 Thus r t traces out a curve in the plane z 4 and x t 2 2 y t 3 2 25 Consequently r t traces out a circle with radius 5 center 2 3 4 keywords vector function space curve circle plane radius center circle 006 part 1 of 2 10 0 points The vector function r t 1 3 cos t i 5 j 2 3 sin t k traces out a circle in 3 space as t varies In which plane does this circle lie keywords 005 10 0 points 1 plane y 5 correct The vector function r t 2 5 sin t i 3 5 cos t j 4 k traces out a circle in 3 space as t varies Determine the radius and center of this circle 2 plane y 5 3 plane x 5 4 plane z 5 1 radius 5 center 2 3 4 correct 5 plane z 5 2 radius 5 center 3 4 2 6 plane x 5 3 radius 5 center 4 3 2 4 radius 3 center 4 3 2 5 radius 3 center 2 3 2 6 radius 3 center 4 3 2 Explanation Writing r t x t i y t j z t k we see that y t 5 for all t Consequently r t traces out a curve in the plane y 5 r t x t i y t j z t k we see that z t 4 for all t while x t 2 5 cos t Explanation Writing y t 3 5 sin t 007 part 2 of 2 10 0 points Determine the radius and center of the circle traced out by r t ha lvh262 Homework 13 1 karakurt 56295 1 radius 3 center 2 1 5 4 r t h 3 4 cos t 2 5 4 sin t i 2 radius 3 center 5 1 2 5 r t h 3 4 sin t 2 4 cos t 5 i 4 3 radius 1 center 5 1 2 6 r t h 3 2 4 cos t 5 4 sin t i correct 4 radius 3 center 1 5 2 correct Explanation If the vector function 5 radius 1 center 5 1 2 r t h x t y t z t i 6 radius 1 center 2 1 2 Explanation Writing r t x t i y t j z t k we see also that x t 1 3 cos t …